Integrand size = 23, antiderivative size = 57 \[ \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {a^2 \sin (c+d x)}{d}-\frac {2 a (a-b) \sin ^3(c+d x)}{3 d}+\frac {(a-b)^2 \sin ^5(c+d x)}{5 d} \] Output:
a^2*sin(d*x+c)/d-2/3*a*(a-b)*sin(d*x+c)^3/d+1/5*(a-b)^2*sin(d*x+c)^5/d
Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {15 a^2 \sin (c+d x)-10 a (a-b) \sin ^3(c+d x)+3 (a-b)^2 \sin ^5(c+d x)}{15 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + b*Tan[c + d*x]^2)^2,x]
Output:
(15*a^2*Sin[c + d*x] - 10*a*(a - b)*Sin[c + d*x]^3 + 3*(a - b)^2*Sin[c + d *x]^5)/(15*d)
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4159, 210, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \tan (c+d x)^2\right )^2}{\sec (c+d x)^5}dx\) |
\(\Big \downarrow \) 4159 |
\(\displaystyle \frac {\int \left (a-(a-b) \sin ^2(c+d x)\right )^2d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle \frac {\int \left ((a-b)^2 \sin ^4(c+d x)-2 a (a-b) \sin ^2(c+d x)+a^2\right )d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \sin (c+d x)+\frac {1}{5} (a-b)^2 \sin ^5(c+d x)-\frac {2}{3} a (a-b) \sin ^3(c+d x)}{d}\) |
Input:
Int[Cos[c + d*x]^5*(a + b*Tan[c + d*x]^2)^2,x]
Output:
(a^2*Sin[c + d*x] - (2*a*(a - b)*Sin[c + d*x]^3)/3 + ((a - b)^2*Sin[c + d* x]^5)/5)/d
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 32.62 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.56
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{5} b^{2}}{5}+2 a b \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )+\frac {a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(89\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{5} b^{2}}{5}+2 a b \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )+\frac {a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(89\) |
risch | \(\frac {5 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {\sin \left (d x +c \right ) a b}{4 d}+\frac {\sin \left (d x +c \right ) b^{2}}{8 d}+\frac {\sin \left (5 d x +5 c \right ) a^{2}}{80 d}-\frac {\sin \left (5 d x +5 c \right ) a b}{40 d}+\frac {\sin \left (5 d x +5 c \right ) b^{2}}{80 d}+\frac {5 \sin \left (3 d x +3 c \right ) a^{2}}{48 d}-\frac {\sin \left (3 d x +3 c \right ) a b}{24 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{16 d}\) | \(143\) |
Input:
int(cos(d*x+c)^5*(a+b*tan(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/5*sin(d*x+c)^5*b^2+2*a*b*(-1/5*cos(d*x+c)^4*sin(d*x+c)+1/15*(2+cos( d*x+c)^2)*sin(d*x+c))+1/5*a^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+ c))
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.25 \[ \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {{\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, a^{2} + a b - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, a^{2} + 4 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")
Output:
1/15*(3*(a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(2*a^2 + a*b - 3*b^2)*cos(d *x + c)^2 + 8*a^2 + 4*a*b + 3*b^2)*sin(d*x + c)/d
\[ \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2} \cos ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**5*(a+b*tan(d*x+c)**2)**2,x)
Output:
Integral((a + b*tan(c + d*x)**2)**2*cos(c + d*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{5} - 10 \, {\left (a^{2} - a b\right )} \sin \left (d x + c\right )^{3} + 15 \, a^{2} \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/15*(3*(a^2 - 2*a*b + b^2)*sin(d*x + c)^5 - 10*(a^2 - a*b)*sin(d*x + c)^3 + 15*a^2*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 2946 vs. \(2 (53) = 106\).
Time = 136.94 (sec) , antiderivative size = 2946, normalized size of antiderivative = 51.68 \[ \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^5*(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")
Output:
-2/15*(15*a^2*tan(1/2*d*x)^10*tan(1/2*c)^9 + 15*a^2*tan(1/2*d*x)^9*tan(1/2 *c)^10 + 20*a^2*tan(1/2*d*x)^10*tan(1/2*c)^7 + 40*a*b*tan(1/2*d*x)^10*tan( 1/2*c)^7 - 75*a^2*tan(1/2*d*x)^9*tan(1/2*c)^8 + 120*a*b*tan(1/2*d*x)^9*tan (1/2*c)^8 - 75*a^2*tan(1/2*d*x)^8*tan(1/2*c)^9 + 120*a*b*tan(1/2*d*x)^8*ta n(1/2*c)^9 + 20*a^2*tan(1/2*d*x)^7*tan(1/2*c)^10 + 40*a*b*tan(1/2*d*x)^7*t an(1/2*c)^10 + 58*a^2*tan(1/2*d*x)^10*tan(1/2*c)^5 - 16*a*b*tan(1/2*d*x)^1 0*tan(1/2*c)^5 + 48*b^2*tan(1/2*d*x)^10*tan(1/2*c)^5 + 150*a^2*tan(1/2*d*x )^9*tan(1/2*c)^6 - 360*a*b*tan(1/2*d*x)^9*tan(1/2*c)^6 + 240*b^2*tan(1/2*d *x)^9*tan(1/2*c)^6 + 700*a^2*tan(1/2*d*x)^8*tan(1/2*c)^7 - 1000*a*b*tan(1/ 2*d*x)^8*tan(1/2*c)^7 + 480*b^2*tan(1/2*d*x)^8*tan(1/2*c)^7 + 700*a^2*tan( 1/2*d*x)^7*tan(1/2*c)^8 - 1000*a*b*tan(1/2*d*x)^7*tan(1/2*c)^8 + 480*b^2*t an(1/2*d*x)^7*tan(1/2*c)^8 + 150*a^2*tan(1/2*d*x)^6*tan(1/2*c)^9 - 360*a*b *tan(1/2*d*x)^6*tan(1/2*c)^9 + 240*b^2*tan(1/2*d*x)^6*tan(1/2*c)^9 + 58*a^ 2*tan(1/2*d*x)^5*tan(1/2*c)^10 - 16*a*b*tan(1/2*d*x)^5*tan(1/2*c)^10 + 48* b^2*tan(1/2*d*x)^5*tan(1/2*c)^10 + 20*a^2*tan(1/2*d*x)^10*tan(1/2*c)^3 + 4 0*a*b*tan(1/2*d*x)^10*tan(1/2*c)^3 - 150*a^2*tan(1/2*d*x)^9*tan(1/2*c)^4 + 360*a*b*tan(1/2*d*x)^9*tan(1/2*c)^4 - 240*b^2*tan(1/2*d*x)^9*tan(1/2*c)^4 - 610*a^2*tan(1/2*d*x)^8*tan(1/2*c)^5 + 2080*a*b*tan(1/2*d*x)^8*tan(1/2*c )^5 - 1200*b^2*tan(1/2*d*x)^8*tan(1/2*c)^5 - 2200*a^2*tan(1/2*d*x)^7*tan(1 /2*c)^6 + 4720*a*b*tan(1/2*d*x)^7*tan(1/2*c)^6 - 2400*b^2*tan(1/2*d*x)^...
Time = 8.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.09 \[ \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {\frac {5\,a^2\,\sin \left (c+d\,x\right )}{8}+\frac {b^2\,\sin \left (c+d\,x\right )}{8}+\frac {5\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{48}+\frac {a^2\,\sin \left (5\,c+5\,d\,x\right )}{80}-\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{16}+\frac {b^2\,\sin \left (5\,c+5\,d\,x\right )}{80}+\frac {a\,b\,\sin \left (c+d\,x\right )}{4}-\frac {a\,b\,\sin \left (3\,c+3\,d\,x\right )}{24}-\frac {a\,b\,\sin \left (5\,c+5\,d\,x\right )}{40}}{d} \] Input:
int(cos(c + d*x)^5*(a + b*tan(c + d*x)^2)^2,x)
Output:
((5*a^2*sin(c + d*x))/8 + (b^2*sin(c + d*x))/8 + (5*a^2*sin(3*c + 3*d*x))/ 48 + (a^2*sin(5*c + 5*d*x))/80 - (b^2*sin(3*c + 3*d*x))/16 + (b^2*sin(5*c + 5*d*x))/80 + (a*b*sin(c + d*x))/4 - (a*b*sin(3*c + 3*d*x))/24 - (a*b*sin (5*c + 5*d*x))/40)/d
Time = 0.21 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {\sin \left (d x +c \right ) \left (3 \sin \left (d x +c \right )^{4} a^{2}-6 \sin \left (d x +c \right )^{4} a b +3 \sin \left (d x +c \right )^{4} b^{2}-10 \sin \left (d x +c \right )^{2} a^{2}+10 \sin \left (d x +c \right )^{2} a b +15 a^{2}\right )}{15 d} \] Input:
int(cos(d*x+c)^5*(a+b*tan(d*x+c)^2)^2,x)
Output:
(sin(c + d*x)*(3*sin(c + d*x)**4*a**2 - 6*sin(c + d*x)**4*a*b + 3*sin(c + d*x)**4*b**2 - 10*sin(c + d*x)**2*a**2 + 10*sin(c + d*x)**2*a*b + 15*a**2) )/(15*d)